Distance — the series
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A distance that is a path through a grid
How far apart two strings are is a shortest-path problem on a grid whose every edge is drawn by the recurrence — and finding a string in a text costs 4,988 character comparisons where measuring how far it is from one costs 96,000.
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A band as wide as the answer
If two strings are close, the optimal route stays near the diagonal and nine cells in ten cannot be on it. A band of three finds the right answer on a pair 300 characters long — and a band of thirty-two is needed before anything can prove it.
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The alignment that fits in one line
Compute the table twice and hold three rows of it. The factor of two is a geometric series and is predicted exactly; measured, it comes down from 2.269 to 2.052 as the strings grow, and the peak is 3(m+1) cells on the nose.
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The row that starts at zero
The same 1,413 cells, filled by the same recurrence in the same order, answer 148 and 0. One line of initialisation decides which question the table was asked, and only one of the two answers is about whether the pattern is there.
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The filter that feeds the table
A self-index answers exact queries and nothing else. Approximate matching needs a table with twenty thousand columns in it. The pigeonhole joins them — cut the pattern into k+1 pieces and at least one occurs exactly, and the index that cannot answer the question decides where to ask it.
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The filter that proposes everything
Seed-and-extend saves two thousand times the work at zero errors and costs more than doing nothing at four. Between them the selectivity falls through the floor, and where it falls is set by two numbers that can be computed before the filter is run — one of which does not contain the length of the text at all.
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The q-grams an error cannot destroy
A pattern of twenty-four characters holds twenty-one four-grams. Two errors can destroy at most eight of them, so any occurrence with two errors still shares thirteen — and a filter that keeps only the windows sharing thirteen proposes 104 of 3,977 and computes 16,744 table cells instead of 96,000.
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The threshold that reaches zero
At q = 5 and four errors on a twenty-four-character pattern the filter demands zero shared q-grams, proposes all 2,977 windows, and computes 986,266 table cells where filling the whole table would have cost 96,000. The failure is arithmetic and is knowable before a character is read.
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The search that spends a budget
A backward search narrows one interval per pattern character. Give it a budget of three errors and it narrows 39,943 of them instead, finds every occurrence the whole table finds, and reads not one character of the text — 177,046 index ranks against 60,000 table cells and zero characters examined.
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The branches an error opens
The tree multiplies by 19.6 for the first error, 12.3 for the second, 10.3 for the third and 8.8 for the fourth. A branching factor of four on a sixteen-character pattern would predict sixty-four, and the gap between sixty-four and eight is the intervals emptying.
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Three savings in three currencies
The same six occurrences, found four ways. The table computes 60,000 cells and reads 60,000 characters. The counting filter computes 6,251 cells and reads 13,022 characters. The seed filter computes 3,325 and reads 550. The index walk computes none, reads none, and performs 18,645 ranks.
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The errors the rest of the pattern needs
Read the pattern left to right in an index of the reversed text and count the points where the interval empties. That count is a lower bound on the errors any alignment of the prefix must contain, it costs 72 rank operations, and it removes 70% of a search tree.
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The branch that cannot reach an answer
Seventy-two rank operations over the pattern remove 27,906 of the 39,957 interval extensions a bounded-error index walk performs — 70% of the tree, at a budget of three. The share grows with the budget, which is what a pruning has to do to be worth its cost.
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A bound that has to be paid for
The pruning removes seventy per cent of a search tree for seventy-two rank operations. It also needs an FM-index of the reversed text — 17,033 bits against the forward index's 17,032 — which doubles the structure whose small size was the entire argument for walking an index.
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The pruning that loses an occurrence
Two versions of the same lower-bound pruning, each one character away from correct. Both return only real occurrences, both return fewer of them, and no check that asks whether the answers are right can tell either from the truth.
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An interval that grows at both ends
A backward search step is two ranks. A bidirectional step is two ranks and the width of the interval for every symbol that sorts before the one being added — 16.5 ranks on four symbols and 138.6 on twenty-six, which is a cost no account of the structure mentions.
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The structure that was supposed to halve
A bidirectional index holds the transform of the text and the transform of its reversal — 35,335 bits each, 70,670 together, exactly twice one index. The deferral that named it hoped it would stop the index doubling. It does not remove the doubling; it reuses it.
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The search that starts in the middle
The same pattern, the same six occurrences, the same index — and 830 interval extensions, or 303, according to which end the search begins at. A pattern cut into three pieces has an error-free one, and only a search with two ends can start there.
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A schedule nobody writes down
A search scheme is valid when its searches between them cover every way the errors can fall — ten cases at two errors, enumerable in a line. A scheme missing one of them finds seven occurrences of eight, all real, at the right error counts, and reports nothing wrong.
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The half that is never asked where
A bidirectional index is two indexes and one interval. One of them is asked for ranks several hundred times a search and for a position never — and the parts that answer "where" are two of the five it is made of.
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An index that cannot locate
Take an FM-index, remove the sampled positions and the bit vector marking them, and refuse every request for a position. What is left still counts, still extends intervals, still runs a whole search — and is a third smaller.
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A sixth of what, exactly
The saving from a counting-only reverse half is 30.5% at one sampled position in four and 14.0% at one in a hundred and twenty-eight. A number quoted without its sampling rate is a number about a setting somebody chose.
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The saving, spent
A bidirectional index whose reverse half cannot locate is a sixth smaller. Give that sixth back to the half that does locate, and the same total size answers a locate five times faster.
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Three savings on one structure
A factor of eleven on an extension, a factor of seven on a branching search, and a sixth of the bits. Applied to one bidirectional index they do not give a factor of seventy-seven, and the reason is that two of the three are the same saving.
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Two factors that do not multiply
Eleven times and seventy-eight times against the same baseline, so an index with both should be eight hundred and sixty. It is seventy-eight, and the shortfall is eleven — the first factor, exactly, because the second operation already contains it.
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The saving that is a loss
An operation that is seventy-eight times cheaper on a branching search costs twice as much on an exact one. It reports every symbol present in order to hand back the one that was asked for, and a search that knows its character needs none of the rest.
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Flat in the budget, and not
One saving is eleven times at every error budget, because it is a property of the alphabet. The other moves between ninety-eight and a hundred and five, because it follows the share of extensions that find nothing. Two savings, two shapes, and neither line crosses the other.
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The ladder, and the rung that spends
Two hundred and fifty thousand bits, then two hundred and twenty-six, then two hundred and eleven. The fourth rung takes the whole saving and buys a four-times denser sampling with it, landing at ninety-six per cent of where it started and locating several times faster.
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Bits and steps on one frame
The size falls from ninety-nine per cent to eighty-six as the sampling thins, and the walk to a sampled position rises from two and a half steps to sixty-four. Neither line is the answer; the answer is a point on the pair.
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The part every saving left alone
A bidirectional index was made smaller three times, one part at a time, and every change left the largest part untouched: the wavelet trees, whose levels are two thirds of the structure. Coding every level in blocks of 63 bits — how many ones a block holds and which arrangement it is — takes that protein index from 210,934 bits to 167,707, a larger step than the two before it together. On English words it takes the same structure from 210,872 bits to 76,054, and an ordered tree, coded, ends up smaller than an unordered Huffman tree, plain. Every rank then reads 1.4 times as many entries, and a smaller superblock buys the reads back for part of the bits.
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Six bits a class
Coding a bidirectional index's wavelet-tree levels in blocks left a part that did not shrink: six bits a block for the block's class, its count of ones, which on English words became a quarter of the levels. Coding the classes by their runs was predicted to halve them. It takes a third off, 0.480 bits a character to 0.323, because two thirds of the blocks are uniform but they come in stretches of about two. One entropy code over the classes does about as well, and on the repetitive collection better. Runs also cut the reads a rank makes by 15%. On random letters runs make the classes larger, and the pointer a variable-length stream needs at every superblock eats most of what an entropy code saves.
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A flag for the blocks that change once
A coded wavelet-tree level stores each mixed 63-bit block as the index of its arrangement, as if every arrangement were equally likely. Coding it instead as the gaps between its value changes was predicted to take a third off the offsets on English words. Alone it adds 4%: a third of the mixed blocks change value once, as predicted, but most of the rest change twenty times or more. A one-bit flag that lets each block take the shorter code takes 21% off, and the levels fall from 1.72 bits a character to 1.45. On the repetitive collection the offsets halve.
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The optimum the flag took away
A bidirectional index's wavelet-tree levels were coded in blocks of 63 bits because a sweep found 63 the best length: shorter blocks pay for more classes, longer ones for offsets that approach the block's own length. Then the classes were coded as runs and each offset given a flag to choose a code of its value changes instead. Lengthening the blocks to 127 bits was predicted to save on English words and the repetitive collection and to change nothing on random letters. It saves on all four texts, and there is no best length short of 1,023 bits on three of them: English words fall from 1.431 bits a character at 63 to 1.304 at 1,023. Under the codings the optimum came from, the same levels cost 3.728 at 1,023. The flag turned a long block's offset into the cost of its change points, which no block length moves, and left the length to be set by the one thing still growing with it: the work of a rank.